MATH383

Overview

Exact solutions to first-order differential equations
Examples

Rewrite a first-order DE as a relationship between infinitesimal increments

y^{'} = \frac{d y}{d x} = f (x, y) \Leftrightarrow M (x, y) d x + N (x, y) d y = 0 . (exact di ff erential)

Compare with $d F (x, y) = F_{x} d x + F_{y} d y$ .

Theorem. If $F : ℝ^{2} \to ℝ$ has continuous partial derivatives $F_{x}, F_{y}$ , then

$F (x, y) = c,$

is an implicit solution of $F_{x} (x, y) d x + F_{y} (x, y) d y = 0$ . ( $F_{x} \equiv M, F_{y} \equiv N$ )

Theorem. If $M, N : ℝ^{2} \to ℝ$ , are continuous with continuous partial derivatives in some open rectangle $R$ , then $M (x, y) d x + N (x, y) d y$ is an exact differential on $R$ if and only if $M_{y} (x, y) = N_{x} (x, y)$ in $R .$

F (x, y) = x^{4} y^{3} + x^{2} y^{5} + 2 x y = c

(%i11)

F: x^4*y^3+x^2*y^5+2*x*y$

(%i12)

diff(F,x);

$(%o12) 2 x y^{5} + 4 x^{3} y^{3} + 2 y$

(%i13)

diff(F,y);

$(%o13) 5 x^{2} y^{4} + 3 x^{4} y^{2} + 2 x$

(2 x y^{5} + 4 x^{3} y^{3} + 2 y) d x + (5 x^{2} y^{4} + 3 x^{4} y^{2} + 2 x) d y = 0 \Rightarrow

\frac{d y}{d x} = - \frac{2 x y^{5} + 4 x^{3} y^{3} + 2 y}{5 x^{2} y^{4} + 3 x^{4} y^{2} + 2 x}, \frac{d x}{d y} = - \frac{5 x^{2} y^{4} + 3 x^{4} y^{2} + 2 x}{2 x y^{5} + 4 x^{3} y^{3} + 2 y} .

Check if the following differential form is exact

3 x^{2} y d x + 4 x^{3} d y = 0 .

Denote: $M (x, y) = 3 x^{2} y, N (x, y) = 4 x^{3}$ . Compute

M_{y} = 3 x^{2}, N_{x} = 12 x^{2} \Rightarrow M_{y} \neq N_{x} \Rightarrow not exact .

Consider $M (x, y) = 4 x^{3} y^{3} + 3 x^{2}, N (x, y) = 3 x^{4} y^{2} + 6 y^{2}$ . Find $F (x, y)$ such that $F_{x} = M$ , $F_{y} = N$ if $M (x, y) d x + N (x, y) d y = 0$ .

Integrate $F_{x} = M$ w.r.t. $x$
$\int F_{x} d x = \int M d x \Rightarrow F - f (y) = \int (4 x^{3} y^{3} + 3 x^{2}) d x = x^{4} y^{3} + x^{2} .$
Replace $F = f (y) + x^{4} y^{3} + x^{3}$ , in $F_{y} = N$
$f^{'} + 3 x^{4} y^{2} = 3 x^{4} y^{2} + 6 y^{2} .$
Integrate to find $f$
$f = 2 y^{3} \Rightarrow F = x^{4} y^{3} + x^{3} + 2 y^{3} .$